Problem: Evaluate $6+\dfrac4a+\dfrac b3$ when $a=4$ and $b=3$.
Solution: Let's substitute $ a= 4$ and $ b={3}$ into the expression. $\phantom{=}6+\dfrac 4{ {a}}+\dfrac{ b}3$ $=6+\dfrac 4{ {4}}+\dfrac{ 3}3$ $=6+1+1$ $=8$